( ) 1.5. Solution: r = mv qb ( 3.2!10 "19 C ) 2.4

Transcription

1 Section 8.4: Motion of Charged Particles in Magnetic Fields Tutorial 1 Practice, page Given: q C; kg; B 2.4 T; v /s Analysis: r v Solution: r v 6.7!10 "27 # kg ) 1.5!10 7 s # kg 3.2!10 "19 C ) 2.4 C ) s r 0.13 Stateent: The radius of the ion s path is Given: q C; kg; B 1.5 T; r 8.0 c Required: v Analysis: r v v r Solution: v r ) !10 "27 kg ) ) 1.60!10 "19 C kg C #s ) v 1.1!10 7 /s Stateent: The speed of the proton is /s. 3. Given: q C; v /s; kg; kg) kg; d Required: B Analysis: r v In a ass spectroeter, the difference between the entry point and the ion detector is 2r. The greater the ass of an ion, the greater the radius. So, the deuteriu ion is detected at 2r fro the entry point, where r is the radius of the path of the hydrogen ion:! d 2r deuteriu! 2r hydrogen Copyright 2012 Nelson Education Ltd. Chapter 8: Magnetic Fields 8.4-1

2 Solution:! d 2r deuteriu! 2r hydrogen 2 deuteriu v! 2 hydrogen v ) 2v 2! hydrogen hydrogen B 2v hydrogen q! d # "10 5 s 1.67 "10!27 kg 1.60 "10!19 C ) ) ) B 8.4 T Stateent: The agnitude of the agnetic field is 8.4 T. 4. a) Since the electric force is up, the balancing agnetic force ust be down. By the righthand rule, the agnetic field should be directed out of the page. b) The agnetic force is F M qvb since the angle is 90. The electric force is F E εq. These forces are equal when the speed is proper: F E F M!q qvb! vb v! B The proper velocity is v! B. c) Since speed only affects the agnetic force, an ion oving too fast will experience a greater agnetic force and be pushed downward. An ion oving too slowly will experience a greater electric force and ove upward. Mini Investigation: Siulating a Mass Spectroeter, page 401 Answers ay vary. Saple answers: A. The ball bearings experience a agnetic force and deflect by different aounts, depending on their asses. This effect is siilar to what happens in a ass spectroeter. B. This activity does not quite odel the function of a ass spectroeter because the bearings do not experience a unifor agnetic force. The force gets stronger at the botto of the rap, and gravity will have a ore significant effect in this siulation than on particles in a ass spectroeter. Copyright 2012 Nelson Education Ltd. Chapter 8: Magnetic Fields 8.4-2

3 Section 8.4 Questions, page The ass spectroeter akes use of the agnetic force on a oving charged particle. Atos are converted into ions and then accelerated into a finely focused bea. The force deflects a particle by an aount depending on its ass and its charge. Electric detectors identify how far the ion travelled in the ass spectroeter. 2. Given: q C) C; U kg; U kg; B 9.5 T; d Required: v Analysis: r v In a ass spectroeter, the difference between the entry point and the ion detector is 2r. The greater the ass of an ion, the greater the radius. So, the U-238 ion is detected at 2r fro the entry point where r is the radius of the path of the U-235 ion:! d 2r U238! 2r U235! d 2r U238! 2r U235 2 U238v! 2 U235v ) 2v U238! U235! d v 2 U238! U235 )! d Solution: v 2 U238! U235 ) kg "10!19 C ) 9.5 C #s ) "10!25 kg! 3.903"10!25 kg ) ) v 1.0 "10 6 /s Stateent: The initial speed of the ions is /s. Copyright 2012 Nelson Education Ltd. Chapter 8: Magnetic Fields 8.4-3

4 3. Given: q C; kg; B T; E k J Analysis: Use E k 1 2 v2 to deterine the speed of the electron; then use r v to deterine the radius of the path. Solution: Deterine the speed of the electron: E k 1 2 v2 v 2E k !10 "19 kg # s # !10 "31 kg ) v 6.954!10 5 /s two extra digits carried) Deterine the radius of the path: r v 9.11!10 "31 # kg ) 6.954!10 5 s # kg 1.60!10 "19 C ) C ) s r 9.34!10 "6 Stateent: The radius of the electron s path is Given: q C; v! /s [E 45 N]; F! 1 is upward; v! /s [up];! F N [W] Required: B Analysis: The upward force in the first situation eans that, by the right-hand rule, the direction of the agnetic field ust be in the x y plane, and soewhere within 180 counterclockwise of E 45 N. The westward force in the second situation eans that, by the right-hand rule, the direction of the agnetic field ust be in the y z plane. The only possible direction that fits both scenarios is north. Use this inforation and F M qvb sin θ to solve for the agnitude of the field. F M qvbsin! F B M qvsin! Copyright 2012 Nelson Education Ltd. Chapter 8: Magnetic Fields 8.4-4

5 Solution: B F M qvsin! 4 "10 #5 kg s 2 4 "10 #9 C) 2 "10 4 ) s * sin90 B 0.5 T Stateent: The agnetic field is 0.5 T [N]. 5. Given: q C; kg; V V; B T Analysis: Use the law of the conservation of energy, E E + E k 0, along with the equations E k 1 2 v2 and! V E E q r v. Solution: Deterine the speed of the electron:! E E +! E k 0 q! V v v2!q! V v!2q! V to deterine the speed of the electron. Then calculate the radius using ) kg # s # 2 C 9.11"10!31 kg )!2!1.60 "10!19 C v "10 6 /s two extra digits carried) Deterine the radius of the path: r v 9.11!10 "31 # kg ) !10 6 s # kg 1.60!10 "19 C ) C ) s r 8.44!10 "4 Stateent: The radius of the path described by the electron is ) Copyright 2012 Nelson Education Ltd. Chapter 8: Magnetic Fields 8.4-5

6 6. a) Given: θ 90 ; v /s; B T Required: ε Analysis: F E F M ; F M qvb sin θ; F E εq F E F M!q qvbsin" Solution:! vbsin90 # 5.0 "10 2 # s kg C)s! 25 N/C Stateent: The strength of the electric field is 25 N/C. b) Given: q C; kg; B T; v /s Analysis: r v Solution: r v " 1.67!10 27 kg ) 5.0!10 2 # s " kg 1.60!10 19 C ) # C s r 1.0!10 4 Stateent: The radius of the proton s path to point P is Copyright 2012 Nelson Education Ltd. Chapter 8: Magnetic Fields 8.4-6